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I Why do we linearise a non linear system

  1. Jun 21, 2016 #1
    I am stuck with the concept of linearising the non-linear system......what i thought was....non-linear systems are defined by differential equations which is difficult to solve, hence we linearise them to make the differential equation look like a algebraic equation....am i right ?......or is there anymore insight that i can get in this topic??
     
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  3. Jun 21, 2016 #2

    Orodruin

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    No, non-linear systems are defined by being non-linear in the function we are trying to solve for.

    No, we linearise to get a differential equation that is linear. Linear differential equations are rather straight forward to solve.

    Typically the solution to the linearised problem is valid in a small region around the solution we linearised the problem around.
     
  4. Jun 21, 2016 #3
    hi orodruin
    1)why is that valid only in small region??thats the point am finding it difficult to understand....am an eletrical engineering student (hence allow me to use electrical example)...
    i have a perception, that is..
    say suppose there is a non-linear system and We are trying to linearise it. we know that for small region dx "slope is constant".....that small region is linear....hence is that the reason why you are saying it is valid only in small region??

    and
    2)other doubt is what is the exact process of linearising... i mean we add dx term right??? and now that model became linear
    3)after linearising....the model equation has say something like this x+dx.....i will process the model accordingly right???


    please correct me if i am wrong
    thanks
    :)
     
  5. Jun 21, 2016 #4

    Orodruin

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    Because if you go further away, the linear description will no longer be good. Compare to approximating a graph with a straight line around some point. Close to the point, you will have a good approximation as long as you use the correct slope for the line. If you go further away, the deviation can become very large.

    We do not add anything, we take a solution and study the deviation from that solution, keeping only terms which are linear in the deviation. The assumption is that the deviation is so small that higher order contributions can be neglected.

    This looks very strange and it is not really clear what you are intending to do.
     
  6. Jun 30, 2016 #5

    jasonRF

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    Orodruin gave great answers. As an electrical engineer let me add an example that may help.

    Consider a transistor. It is a fundamentally nonlinear device - just look at the I-V curves that you get out of one. When you are doing "small signal analysis" of a transistor circuit you start with the DC bias point, which sets the mode of operation (forward active, saturation, cutoff, or reverse active). The "small signal analysis" is actually a linearization of this nonlinear circuit about that DC operating point. It is useful because for these small signals you can use linear circuit elements to represent the transistor and gain both a qualitative and quantitative understanding of how the transistor circuit works. However, once your signal get too large and, say, you start to saturate a transistor amplifier, then clearly the "small signal analysis" no longer holds. That is, the linear description fails since you are too far away from that DC operating point that you linearized about.

    Hopefully that helps.

    Jason
     
  7. Jul 17, 2016 #6

    epenguin

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    It is just a standard part of the qualitative analysis of nonlinear differential equations. Usually you cannot get the 'closed' solutions in terms of familiar functions that you can for linear equations. However you can generally get a good qualitative idea.

    Usually the first step is to find the 'stationary points', that is the points where all the first derivatives with respect to time are zero. There is usually one, but there may be more. Their existence may depend on the values of the parameters of the equation. At certain values of these, new stationary points may come into existence or vanish corresponding to a qualitative change in the behaviour which is a hallmark of non-linearity.

    Then usually you transform the equation so as to make a stationary point a zero point (that is for example for two variables the .point (0, 0). Then ignore the nonlinear terms and you have an approximation to the behaviour near to (0, 0). The most important thing is whether these points are attractive or repulsive, i.e. eigenvalues negative or positive, just like you have already learnt and maybe forgotten https://www.physicsforums.com/cid:F4D2AA33-AD19-4BFB-84AB-4FF46835381F@home [Broken] for linear differential equations. (Can be attractive in one dimension, repulsive in another.) If the point is attractive/repulsing locally, in the linear approximation, so will it be for the full equation for some way out at least. In the case of linear differential equations there is a single stationary point which is either attractive or responsive over infinite distance, or over the physically possible distance. Whereas the attractive or repulsion range of non-linear equation may be only finite. (A stationary point may be surrounded by a limit cycle for example, but the linear analysis by itself does not tell you that one way or the other.)

    The way you build up a qualitative picture of the behaviour of a non-linear differential equation combining this analysis and other tricks is quite reminiscent of curve sketching - the way you build up an understanding of the shape of a curve from bits of calculation in elementary calculus. To get the exact solutions, which as I said most often cannot be done analytically, you use a numerical DE solving routine or app. SP analysis however is very useful in telling you what are the parameter ranges in which the nonlinear behaviour of interest occurs - otherwise you would have to explore a perhaps large parameter range inefficiently by computer experiment.
     
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  8. Jul 17, 2016 #7

    FactChecker

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    I think you mean this: If you linearize the problem at a point x0, then the linearized model equation has variable dx where x = x0+dx.

    The linearized problem has the constants at the point x0 and the linear slopes are multiplied by dx.
     
  9. Jul 17, 2016 #8

    Stephen Tashi

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    If you are using the standard terminology when you say "linearise them" then you've gotten some answers. But let's make sure you aren't talking about writing the "characteristic equation" for a differential equation. That associates an algebraic equation with a differential equation, but it isn't what we call "linearising" it.
     
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